Problem 42 - Chemistry and Chemical Reactivity

Question

Phosphorus occurs in nature as a single isotope, phosphorus- \(31 .\) Neutron irradiation of phosphorus- \(31\) produces phosphorus- \(32,\) a \(\beta\) emitter with a half-life of 14.28 days. Assume you have a sample containing phosphorus- 32 that has a rate of decay of \(3.2 \times 10^{6} \mathrm{dpm.}\) Draw a graph showing disintegrations per minute as a function of time during a period of one year.

Step-by-Step Solution

Answer

Graph the decay using the formula \(N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{14.28}}\) for a year.

Step 1: Understand the Problem

We're given phosphorus-32, a radioactive isotope with a half-life of 14.28 days, and an initial decay rate of \(3.2 \times 10^{6} \mathrm{dpm}\). We need to graph the decay rate over one year.

Step 2: Determine Key Values

Since the initial rate of decay is \(3.2 \times 10^{6} \mathrm{dpm}\), we know this represents the number of disintegrations per minute at the starting time (\(t=0\)). The half-life is 14.28 days.

Step 3: Use the Half-Life Formula

The decay of a radioactive substance can be described using the formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T}} \]where \(N(t)\) is the number of disintegrations per minute at time \(t\), \(N_0\) is the initial number of disintegrations per minute, and \(T\) is the half-life.

Step 4: Calculate Rate of Decay Over Time

Using the formula from Step 3, substitute \(N_0 = 3.2 \times 10^{6} \), \(T = 14.28\) days, and keep \(t\) as a variable in days. This will give us an equation to calculate \(N(t)\) at any given time.

Step 5: Set Up the Graph

Make a graph with time on the x-axis (in days, up to 365 days for one year), and the decay rate on the y-axis. Using the formula from Step 4, calculate the rate at different time points (e.g., at each half-life or every 30-60 days).

Step 6: Plot and Connect Data Points

For each calculated point in Step 5, plot the corresponding point on the graph. Connect these points smoothly since the decay is exponential.

Step 7: Verify and Analyze

Ensure that the graph decreases exponentially, reflecting the nature of radioactive decay, and verify that at each half-life (multiples of 14.28 days), the rate should be about half of the previous value.

Key Concepts

Phosphorus Isotopes

Phosphorus, a vital element for all living organisms, primarily exists in nature as the isotope phosphorus-31. Isotopes are variants of a particular chemical element that differ in neutron number, but share the same number of protons. In this context, we deal with phosphorus-32, a radioactive isotope created by neutron irradiation of phosphorus-31. While phosphorus-31 is stable and non-radioactive, phosphorus-32 undergoes radioactive decay. This quality makes phosphorus-32 useful in various scientific applications, such as tracing and research at the molecular level.

Half-Life

The concept of half-life is crucial in understanding radioactive decay. It refers to the time required for a quantity to reduce to half its initial value. In the case of phosphorus-32, its half-life is 14.28 days. This means that every 14.28 days, the activity of phosphorus-32 will decrease to half its initial value. This predictable pattern helps in calculating how quickly a radioactive substance will decay over time. For example, if you start with a rate of decay of \(3.2 \times 10^6\) disintegrations per minute (dpm), after 14.28 days, this rate will reduce to \(1.6 \times 10^6\) dpm.

Exponential Decay

Radioactive substances decay in a characteristic manner known as exponential decay. In exponential decay, the rate of decrease is proportional to the current amount, resulting in a smooth, continuous reduction over time. The mathematical representation of this process is given by \(N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T}}\). Here, \(N(t)\) is the remaining amount at time \(t\), \(N_0\) is the initial quantity, and \(T\) is the half-life. The graph of this equation forms a downward-sloping curve. As time progresses, the rate decreases exponentially, reflecting how quickly a radioactive substance loses activity over time.

Beta Emission

Beta emission is one of the types of radioactive decay processes. Specifically, phosphorus-32 undergoes beta decay. In beta emission, a neutron in the nucleus is transformed into a proton, releasing a beta particle, which is an electron. This process alters the atomic number and transmutes one element into another, but doesn't change the atomic mass significantly. For phosphorus-32, this emission turns it into sulfur-32, a stable element. Understanding beta emission is key because it helps illustrate how radioactive isotopes like phosphorus-32 reduce their radioactivity over time, transitioning into stable configurations.